Let $a,b,c>0$. Prove $$\sqrt{\frac{a}{b+3c}}+\sqrt{\frac{b}{c+3a}}+\sqrt{\frac{c}{a+3b}}\ge \frac{3}{2}$$


Holder: $A^2\cdot Σ_{cyc}\left(a^2\left(b+3c\right)\right)\ge \left(a+b+c\right)^3$

So we need to prove $4\left(a+b+c\right)^3\ge 9Σ_{cyc}\left(a^2\left(b+3c\right)\right)$

$\Leftrightarrow 4a^3+4b^3+4c^3+3a^2b+3b^2c+3c^2a+24abc\ge 15ab^2+15bc^2+15a^2c$

Then i don't know how to solve it. Help me !

  • $\begingroup$ Is any condition else given? $\endgroup$ – Dr. Sonnhard Graubner Apr 20 '18 at 13:17
  • $\begingroup$ @Dr. Sonnhard Graubner: no $\endgroup$ – Word Shallow Apr 20 '18 at 13:33

Your trying gets a wrong inequality.

Try $c=0$ and $a=b=1$.

Let $\frac{a}{b+3c}=\frac{x^2}{4}$, $\frac{b}{c+3a}=\frac{y}{4}$ and $\frac{c}{a+3b}=\frac{z^2}{4}$, where $x$, $y$ and $z$ are non-negative numbers.

Hence, the system $$\begin{array}{l}4a-x^2b-3x^2c=0\\-3y^2a+4b-y^2c=0\\ -z^2a-3z^2b+4c=0\\ \end{array}$$ has infinitely many solutions $(a,b,c)$,

Hence $$ \|(\begin{array}{ccc}4& -x^2 & -3x^2\\ -3y^2 & 4 & -y^2\\ -z^2 & -3z^2 & 4\end{array})\| = 0 ,$$ which gives $$3(x^2y^2+x^2z^2+y^2z^2)+7x^2y^2z^2=16$$ and we need to prove that $$x+y+z\geq3.$$ Let $x+y+z<3$, $x=kp$, $y=kq$ and $z=kr$ such that $k>0$ and $p+q+r=3$.

Hence, $k(p+q+r)<3$, which gives $0<k<1$.

Thus, $$16=3(x^2y^2+x^2z^2+y^2z^2)+7x^2y^2z^2=3k^4(p^2q^2+p^2r^2+q^2r^2)+7k^6p^2q^2r^2<$$ $$<3(p^2q^2+p^2r^2+q^2r^2)+7p^2q^2r^2.$$ But it's a contradiction because we'll prove now that $$16\geq3(p^2q^2+p^2r^2+q^2r^2)+7p^2q^2r^2$$ for all non-negatives $p$, $q$ and $r$ such that $p+q+r=3$.

Indeed, let $p+q+r=3u$, $pq+pr+qr=3v^2$ and $pqr=w^3$.

Hence, $16\geq3(p^2q^2+p^2r^2+q^2r^2)+7p^2q^2r^2\Leftrightarrow f(w^3)\leq0$, where $f$ is a convex function.

Hence, by $uvw$ it remains to check two cases.

  1. $w^3=0$.

Let $r=0$.

We need to prove that $$16\geq3p^2q^2,$$ which is true because $$p^2q^2\leq\frac{(p+q)^4}{16}=\frac{81}{16}<\frac{16}{3}.$$

  1. $q=r$.

After homogenization we can assume $q=r=1$ and it remains to prove that $$\frac{16(p+2)^6}{729}\geq7p^2+\frac{(2p^2+1)(p+2)^2}{3}$$ or $(p-1)^2(8p^4+112p^3+453p^2+1102p+26)\geq0$, which is obvious.


  • $\begingroup$ But $c>0$ why do you choose $c=0$ ? $\endgroup$ – Word Shallow Jun 9 '18 at 13:00
  • $\begingroup$ Try $c=0.0001$ and $a=b=1$. $\endgroup$ – Michael Rozenberg Jun 9 '18 at 14:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.